Question

Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of C2H4 is 52.26 kJ/mol, CO2 is -393.509 kJ/mol, and H2O is -241.818 kJ.) C2H4 (g) + 3O2(g) 2CO2 (g) + 2H2O(g)

ΔHrxn =
The reaction is .

Answer 1

Answer: -355.642

See picture for explanation

Answer 2

Answer: The enthalpy change of the reaction is -1322.91 kJ

Step-by-step explanation:

The chemical equation for the combustion of propane follows:

`C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(2* \Delta H^o_f_((CO_2(g))))+(2* \Delta H^o_f_((H_2O(g))))]-[(1* \Delta H^o_f_((C_2H_4(g))))+(3* \Delta H^o_f_((O_2(g))))]`

We are given:

`\Delta H^o_f_((H_2O(g)))=-241.818kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((CO_2(g)))=-393.509kJ/mol\\\Delta H^o_f_((C_2H_4(g)))=52.26kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(2* (-393.509))+(2* (-241.818))]-[(1* (52.26))+(3* (0))]\\\\\Delta H^o_(rxn)=-1322.91kJ`

Hence, the enthalpy change of the reaction is -1322.91 kJ